Find the area of a quadrilateral ABCD in which AB=3 cm, BC=4 cm, CD=4 cm, DA=5 cm and AC=5 c

2S=5+4+3 ⇒ S=6

Area of ΔABC

=

s(s−a)(s−b)(s−c)

=

15(15−9)(15−8)(15−13)

=

6(1)(2)(3)

=

36

=6

⇒For area of ΔADC,

⇒2S=5+4+5

⇒ S=7cm

⇒Area of ΔADC

=

7(7−5)(7−5)(7−4)

=

7×2×2×3

=2

21

=9.16m

2

⇒Area of quad ABCD= Area of (△ABC+△ADC)

=6+9.16

=15.16 cm

2

≈15.2 cm

2